Author Topic: 4U Maths Question Thread (Read 816019 times) Tweet Share

RuiAce · « **Reply #1125 on:** April 27, 2017, 07:59:02 pm »

Quote from: beau77bro on April 27, 2017, 07:46:08 pm

(Image removed from quote.)

This is an absolute monster question. I think atleast ahahaha any help is ridiculously appreciated

$\text{The series is just}\\ 1+z+z^2+\dots+z^n = \frac{1-z^{n+1}}{1-z}$
$\text{If }z=x(\cos x + i\sin x)\text{ then by de Moivre's formula}\\ 1+z+\dots+z^n = \frac{1-x^{n+1}(\cos [(n+1)x] + i \sin [(n+1)x])}{1-x(\cos x + i\sin x)}$
I don't see how that can be simplified any further and neither does Wolfram

legorgo18 · « **Reply #1126 on:** April 27, 2017, 09:54:07 pm »

Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

RuiAce · « **Reply #1127 on:** April 27, 2017, 11:30:09 pm »

Quote from: legorgo18 on April 27, 2017, 09:54:07 pm

Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

$\text{Working backwards on scrap paper}\\ \begin{align*}\frac1x +\frac1y& \ge \frac{4}{x+y}\\ y+x&\ge \frac{4xy}{x+y}\\ (x+y)^2=x^2+2xy+y^2&\ge 4xy\\ x^2-2xy+y^2&\ge 0\\ (x-y)^2&\ge 0\end{align*}$
$\text{So write out the exact same thing in the right order}\\ \text{by writing the above 'proof' in reverse}\\ \text{i.e. start from }(x-y)^2\ge 0$
____________________

I have no idea what they teach there because the Cauchy-Schwarz inequality isn't even in the HSC...
I do not see how this benefits students when they can't even use it in the exam.
$\text{To apply the result }\left(\sum_{i=1}^3 a_i b_i\right)^2 \le \left(\sum_{i=1}^3 a_i^2\right)\left(\sum_{i=1}^3 b_i^2\right)\\ \text{Let }\{a_i\}_{i=1}^3\text{ be the sequence }a_1=\sqrt{2a},a_2\sqrt{2b},a_3=\sqrt{2c}\\ \text{and let }\{b_i\}_{i=1}^3\text{ be the sequence }b_1=\sqrt{bc},b_2=\sqrt{ac},b_3=\sqrt{ab}$
$\text{Then,}\\ \begin{align*}(\sqrt{2a}\sqrt{bc}+\sqrt{2b}\sqrt{ac}+\sqrt{2c}\sqrt{ab})^2 &\le (2a+2b+2c)(bc+ac+ab)\\ (3\sqrt{2abc})^2&\le 2(abc+a^2c+a^2b+b^2c+abc+b^2a+bc^2+ac^2+abc)\\ 9abc &\le ab(a+b)+ac(a+c)+bc(b+c)+3abc\\ 6abc&\le ab(a+b)+ac(a+c)+bc(b+c)\end{align*}\\ \text{hence proving the second half}$
Remarks: A lot of algebra was rushed. You may consider working through the algebra more slowly.
For the linear algebra version |a.b| ≤ ||a|| ||b||, simply treat the terms of the sequence as the components of the vectors instead.

Haven't figured out how C-S can be used for the first half yet

RuiAce · « **Reply #1128 on:** April 28, 2017, 03:11:07 pm »

Quote from: legorgo18 on April 27, 2017, 09:54:07 pm

Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

$\text{Upon further investigation, I've found that what you call 'Cauchy's inequality}\\ \text{is actually the AM-GM inequality}\\ \text{which, whilst still not assumable, is much more fit for use.}$
I have absolutely no clue why it's called that though - that's wrong to me.
$\text{For 1), if one is allowed to assume the two-variable AM-GM inequality}\\ \begin{align*}\frac{x+y}{2} &\ge \sqrt{xy}\\ (x+y)^2&\ge 4xy\\ {x+y}&\ge \frac{4xy}{x+y}\\ \frac{1}y+\frac1x&\ge \frac{4}{x+y}\end{align*}$
This means I'll have to look at 2) properly later to determine what to substitute into the 3-variable AM-GM inequality.

Edit: Found a convoluted proof for part 2
$\text{Note that}\boxed{\frac{x}{y}+\frac{y}{x} \ge 2}\\ \text{The proof of this is left as an exercise - it follows from the 2-variable AM-GM}$
$\text{Then, applying this }\textbf{three times and THEN multiplying}\\ \begin{align*}\left(\frac{x}y+\frac{y}x\right)\left(\frac{x}z+\frac{z}x\right)\left(\frac{y}z+\frac{z}y\right)&\ge 8\\ \left(\frac{x^2+y^2}{xy}\right)\left(\frac{x^2+z^2}{xz}\right)\left(\frac{y^2+z^2}{yz}\right)&\ge 8\\ \sqrt[3]{\frac{(x^2+y^2)(x^2+z^2)(y^2+z^2)}{x^2y^2z^2}}\ge 2\end{align*}$
$\text{Let:}\\ \begin{align*}x^2&=a\\y^2&=b\\z^2&=c\end{align*}$
$\text{Then, }\begin{align*}\sqrt[3]{\frac{(a+b)(a+c)(b+c)}{abc}}&\ge 2\\\sqrt[3]{a^2b^2c^2(a+b)(a+c)(b+c)}&\ge 2abc \end{align*}$
$\text{The AM-GM inequality for three variables is}\\ \boxed{x+y+z\ge 3\sqrt[3]{xyz}}\\ \text{Let:}\\ \begin{align*}x&=ab(a+b)\\ y&=bc(b+c)\\ z&=ac(a+c)\end{align*}$
$\text{Then, combining AM-GM with the lemma above}\\ \begin{align*}ab(a+b)+bc(b+c)+ac(a+c)&\ge 3\sqrt[3]{a^2b^2c^2(a+b)(a+c)(b+c)}\\ &\ge 6abc\end{align*}$

beau77bro · « **Reply #1129 on:** April 29, 2017, 01:11:25 pm »

Quote from: RuiAce on April 27, 2017, 07:59:02 pm

$\text{The series is just}\\ 1+z+z^2+\dots+z^n = \frac{1-z^{n+1}}{1-z}$
$\text{If }z=x(\cos x + i\sin x)\text{ then by de Moivre's formula}\\ 1+z+\dots+z^n = \frac{1-x^{n+1}(\cos [(n+1)x] + i \sin [(n+1)x])}{1-x(\cos x + i\sin x)}$
I don't see how that can be simplified any further and neither does Wolfram

Omg THANKYOU tho I don't know series. The answer is this if it helps. Sorry for such a random Q and it probably would've helped a lot if I'd shown where u actually need to get to sorrryy

Maybe this will help? All g if it doesn't just stumped

RuiAce · « **Reply #1130 on:** April 29, 2017, 04:35:20 pm »

Quote from: beau77bro on April 29, 2017, 01:11:25 pm

Omg THANKYOU tho I don't know series. The answer is this if it helps. Sorry for such a random Q and it probably would've helped a lot if I'd shown where u actually need to get to sorrryy(Image removed from quote.)

Maybe this will help? All g if it doesn't just stumped

$\text{Series is an important part of 2U and at times you will definitely need it for complex numbers}$
$\text{The given answers choose to make the denominator real}\\ \text{Hence, apply the typical conjugate tricks}\\ \text{Note that the remainder of this solution is nothing more than stupidly messy algebra.}$
$\begin{align*}&\quad \frac{1-x^{n+1}\cos (n+1)x - i x^{n+1}\sin (n+1)x}{1-x\cos x - ix\sin x}\\ &= \frac{1-x^{n+1}\cos (n+1)x - i x^{n+1}\sin (n+1)x}{(1-x\cos x) - (ix\sin x)}\times\frac{1-x\cos x+ix\sin x}{(1-x\cos x)+i(x\sin x)}\end{align*}$
$\text{At this point things just get messy so I will just do the numerator and denominator separately}\\ \text{The denominator is relatively easy}\\$
\begin{align*}&\quad \left[(1-x\cos x)-i(x\sin x)\right][(1-x\cos x)+i(x\sin x)]\\ &= (1-x\cos x)^2+(x\sin x)^2\\ &= 1-2x\cos x + x^2(\cos^2x+\sin^2x)\\ &= 1-2x\cos x + x^2\end{align*}
$\text{As for the numerator, recall the following results}\\ \begin{align*} \text{ cis } \alpha \times \text{ cis } \beta&= \text{ cis } (\alpha+\beta)\\ \text{ cis } (-\theta)&=\cos \theta - i \sin \theta\end{align*}$
Cis notation is ew.
$\begin{align*}&\quad \left(1-x^{n+1} \text{ cis } [(n+1)x]\right)\left(1-x \text{ cis } (-x)\right)\\&= 1 - x \text{ cis } (-x) - x^{n+1} \text{ cis } [(n+1)x] + x^{n+2} \text{ cis } nx\end{align*}$
$\text{Since the solutions split up the real and imaginary parts}\\ \begin{align*}a&=1 - x\cos (-x) - x^{n+1}\cos [(n+1)x] + x^{n+2}\cos nx\end{align*}\\ \text{then using }\cos(-x) = \cos x\text{ we are done}$
I might've made an algebra mistake because I don't quite agree with what they got for b.

Kle123 · « **Reply #1131 on:** April 30, 2017, 10:43:53 am »

HEY, could anyone help me out with this integration? THANK YOU. I edited the red minus in(forgot to rewrite it after erasing it). The answer is also attached.

RuiAce · « **Reply #1132 on:** April 30, 2017, 11:06:25 am »

Quote from: Kle123 on April 30, 2017, 10:43:53 am

HEY, could anyone help me out with this integration? THANK YOU. I edited the red minus in(forgot to rewrite it after erasing it). The answer is also attached.

The method of computing your integral can be found in post #1040. In your scenario, take R = 6

Kle123 · « **Reply #1133 on:** April 30, 2017, 01:08:47 pm »

Sorry Rui, I dont understand how you got from the first to second step and how the integration of the first part of the 2nd step is equal to the first part of the 3rd step. Could you explain? Thanks

RuiAce · « **Reply #1134 on:** April 30, 2017, 01:10:48 pm »

Quote from: Kle123 on April 30, 2017, 01:08:47 pm

Sorry Rui, I dont understand how you got from the first to second step and how the integration of the first part of the 2nd step is equal to the first part of the 3rd step. Could you explain? Thanks

$\text{First step is just multiplying top and bottom by }\sqrt{x}$
$\text{You should be able to do this integral quite easily}\\ \int \frac{R-x}{\sqrt{2R-x^2}}dx=-\sqrt{2R-x^2}+C\\ \text{If it is absolutely unclear, use the substitution }u=2R-x^2$
$\text{It works like }\int \frac{x}{\sqrt{x^2-1}}dx=\sqrt{x^2-1}+C$

Kle123 · « **Reply #1135 on:** April 30, 2017, 01:40:38 pm »

Sorry. Before I meant i didn't know how to get from the first line to the second line on the right hand side. Also instead i tried differentiating it to check the integration but i got different.

RuiAce · « **Reply #1136 on:** April 30, 2017, 01:52:41 pm »

Quote from: Kle123 on April 30, 2017, 01:40:38 pm

Sorry. Before I meant i didn't know how to get from the first line to the second line on the right hand side. Also instead i tried differentiating it to check the integration but i got different.

Oh my bad, this whole time there were typos floating around.
$\begin{align*}\int \frac{x}{\sqrt{2Rx-x^2}}dx &= -\int \frac{-x}{\sqrt{2Rx-x^2}}dx\\&=-\int \left( \frac{R-x-R}{\sqrt{2Rx-x^2}}\right)dx\\ &=\int \left(\frac{R-x}{\sqrt{2Rx-x^2}}+ \frac{-R}{\sqrt{2Rx-x^2}} \right)dx\end{align*}$
Given the typo's I'm not too sure how my original answer came out nicely (unless it didn't).

That being said, these are 4U integration tricks that you are expected to know.

Kle123 · « **Reply #1137 on:** April 30, 2017, 02:25:59 pm »

Quote from: RuiAce on April 30, 2017, 01:52:41 pm

That being said, these are 4U integration tricks that you are expected to know.

I see. Thanks for the help Rui!

chelseam · « **Reply #1138 on:** May 01, 2017, 11:45:17 pm »

Hi! Could someone please help me with this question? The exercise is about using the substitution t=tanθ/2. Thank you

RuiAce · « **Reply #1139 on:** May 02, 2017, 06:55:40 am »

Quote from: chelseam on May 01, 2017, 11:45:17 pm

Hi! Could someone please help me with this question? The exercise is about using the substitution t=tanθ/2. Thank you

What textbook is this? That substitution is a horrible idea; u = sin(x) would be a much better substitution.

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